package leetcode.editor.cn;
//给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并
//返回其根节点。 
//
// 
//
// 示例 1: 
//
// 
//输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
//输出: [3,9,20,null,null,15,7]
// 
//
// 示例 2: 
//
// 
//输入: preorder = [-1], inorder = [-1]
//输出: [-1]
// 
//
// 
//
// 提示: 
//
// 
// 1 <= preorder.length <= 3000 
// inorder.length == preorder.length 
// -3000 <= preorder[i], inorder[i] <= 3000 
// preorder 和 inorder 均 无重复 元素 
// inorder 均出现在 preorder 
// preorder 保证 为二叉树的前序遍历序列 
// inorder 保证 为二叉树的中序遍历序列 
// 
// Related Topics 树 数组 哈希表 分治 二叉树 
// 👍 1929 👎 0


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import java.util.HashMap;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution83 {
    // 存储 inorder 中值到索引的映射
    HashMap<Integer, Integer> valToIndex = new HashMap<>();

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        for (int i = 0; i < inorder.length; i++) {
            valToIndex.put(inorder[i], i);
        }

        // 根据函数定义，用 preorder 和 inorder 构造二叉树
        return build(preorder, 0, preorder.length - 1,
                inorder, 0, inorder.length - 1);
    }

    /**
     * build 函数的定义：
     * 若前序遍历数组为 preorder[preStart..preEnd]，[root,leftNode,....,rightNode,....]
     * 中序遍历数组为 inorder[inStart..inEnd]，[leftNode,....,root,rightNode,....]
     * 构造二叉树，返回该二叉树的根节点
     *
     * @param preorder
     * @param preLeft
     * @param preRight
     * @param inorder
     * @param inLeft
     * @param inRight
     * @return
     */
    private TreeNode build(int[] preorder, int preLeft, int preRight, int[] inorder, int inLeft, int inRight) {
        if (preLeft <= preRight) {
            int rootVal = preorder[preLeft];
            TreeNode root = new TreeNode(rootVal);
            // 避免 for 循环寻找 rootVal
            int index = valToIndex.get(rootVal);
            int leftSize = index - inLeft;
            root.left = build(preorder, preLeft + 1, preLeft + leftSize,
                    inorder, inLeft, inLeft + leftSize - 1);

            root.right = build(preorder, preLeft + leftSize + 1, preRight,
                    inorder, inLeft + leftSize + 1, inRight);
            return root;
        }
        return null;
    }

    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode() {
        }

        TreeNode(int val) {
            this.val = val;
        }

        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }
}
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